Solving Quadratic Equations by the Quadratic Formula
July 16, 2025
A quadratic equation is a polynomial equation of the form:
where a, \, b \, \text{ and } c are real numbers and a \neq0.
Before we can solve any quadratic. it helps to have a general tool that works for all cases.
Any quadratic equation ax^2+bx+c=0 has solutions given by:
To ensure you apply it correct every time, follow these simple guidelines.
Follow these steps to use the quadratic formula effectively:
- Standard Form. Ensure the equation is written as ax^2+bx+c=0.
- Identify Coefficients. Read off a, \, b, \, \text{ and } c then substitute into
- Compute and simplify. Evaluate the discriminant \Delta = b^2 - 4ac, simplify the square root and solve for the two values of x.
Once you have your solutions, the discriminant tells you what kind of roots to expect.
The sign of the discriminant \Delta = b^2 - 4ac determines the nature if the solutions:
- \Delta > 0 : Two distinct real roots.
- \Delta = 0: One real (repeated) root.
- \Delta < 0: Two complex roots.
Illustrative Example #1
Find the solution of the equation x^2-5x+6=0 using the quadratic formula.
Step 1: Standard Form
Notice that x^2-5x+6=0 is already in standard form.
Step 2: Identify and Substitute
Here a=1, \, b=-5,\, c=6. Thus, by the quadratic formula, we obtain:
Since the discriminant \Delta = b^2 - 4ac = 1 > 0, there are two distinct real roots.
Step 3: Simplify
Hence, the solutions to x^2-5x+6=0 are:
Illustrative Example #2
Find the solutions of the equation (x-3)(x+4)=42 using the quadraitc formula.
Step 1: Standard Form
We need to algebraically manipulate to its standard form first:
Step 2: Identify and Substitute
Here a = 1, \, b = 1,\text{ and } c = −54. Substitute into the quadratic formula:
Since the discriminant ∆ = 217 > 0, there are two distinct real roots.
Step 3: Simplify
Therefore, the solutions are:
Illustrative Example #3
Find the solutions of the equation \displaystyle \bm{\frac{1}{2}x^2 - \frac{3}{4}x - \frac{5}{2} = \frac{3}{4}} using the quadratic formula.
Step 1: Standard Form
We need to algebraically manipulate\displaystyle \bm{\frac{1}{2}x^2 - \frac{3}{4}x - \frac{5}{2} = \frac{3}{4}} to its standard form first.
Step 2: Identify and Substitute
Here a = \dfrac{1}{2}, \, b = -\dfrac{3}{4}, \text{ and } c = -\dfrac{13}{4}. Substitute into the quadratic formula:
Since the discriminant \Delta = \dfrac{113}{16} > 0 there are two distinct real roots.
Step 3: Simplify
Therefore, the solutions are:
Illustrative Example #4
Find the solutions of the equation x^2-6x+9=0 using the quadratic formula.
Step 1: Standard Form
The equation is already in standard form: x^2-6x+9=0.
Step 2: Identify and Substitute
Here a = 1, \, b = −6, \text{ and } c = 9. Substitute into x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}:
Since the discriminant \Delta = 36 - 36 = 0, there is one real (repeated) root.
Step 3: Simplify
Therefore, the solution is:
Illustrative Example #5
Find the solutions of the equation (x − 4)(x − 2) = −1 using the quadratic formula.
Step 1: Standard Form
We need to algebraically manipulate (x − 4)(x − 2) = −1 to its standard form first.
Step 2: Identify and Substitute
Here a = 1, \,b = −6, \text { and }c = 9. Substitute into the quadratic formula:
Since the discriminant \Delta = 0, there is one real (repeated) root.
Step 3: Simplify
Therefore, the solution is:
In order to solve equations with negative discriminant, we introduce the imaginary unit:
This definition extends the real numbers to the complex numbers, allowing us to write \sqrt{b^2 - 4ac} even when b^2 - 4ac < 0.
Illustrative Example #6
Find the solutions of the equation x^2+2x+5=0 using the quadratic formula.
Step 1: Standard Form
Notice that x^2+2x+5=0 is already in standard form.
Step 2: Identify and Substitute
Here a = 1, \, b = 2, \text{ and }c = 5. Substitute into the quadratic formula:
Since the discriminant \Delta = 4 - 20 = -16 < 0, there are two complex conjugate roots.
Step 3: Simplify
Therefore, the solutions are:
Illustrative Example #7
Find the solutions of the equation (x + 2)(x − 1) = −3 using the quadratic formula.
Step 1: Standard Form
We need to FOIL and then rewrite (x + 2)(x − 1) = −3 in standard form:
Step 2: Identify and Substitute
Here a = 1, \, b = 1, \text{ and }c = 1. Substitute into x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}:
Since the discriminant \Delta = -3 < 0, there are two complex conjugate roots.
Step 3: Simplify
Therefore, the solutions are: